Infinite series - Inverse trigonometry

[Saitama]

Homework Statement

The sum of the infinite terms of the series
\text{arccot}\left(1^2+\frac{3}{4}\right)+\text{arccot}\left(2^2+\frac{3}{4}\right)+\text{arccot}\left(3^2+\frac{3}{4}\right)+...
is equal to
A)arctan(1)
B)arctan(2)
C)arctan(3)
D)arctan(4)

Ans: B

Homework Equations

The Attempt at a Solution

The given series can be written as
$$\sum_{r=1}^{\infty} \text{arccot}\left(r^2+\frac{3}{4}\right)$$
Simplifying the term inside the summation, I rewrite it as
$$\arctan\left(\frac{4}{4r^2+3}\right)$$
I don't see how to proceed further. I am thinking of converting the above in the form $\arctan\left(\frac{x-y}{1+xy}\right)$, rewrite it in the form $\arctan(x)-\arctan(y)$ and perform the summation but I can't see a way to do this.

Any help is appreciated. Thanks!

 

[vela]

I haven't worked it out, but I think you're on the right track. Try rewriting it as
$$r^2 + \frac{3}{4} = \left(r^2 - \frac{1}{4}\right) + 1 = \left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right) + 1$$

 

[Saitama]

I haven't worked it out, but I think you're on the right track. Try rewriting it as
$$r^2 + \frac{3}{4} = \left(r^2 - \frac{1}{4}\right) + 1 = \left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right) + 1$$

Wow. That's great. Thanks a lot vela!

I did try rewriting 3/4 as 1-1/4 but then I didn't notice that it could simplified to (r-1/2)(r+1/2).

 

[verty]

This is one of the most sadistic questions I've seen. Where are you getting these questions?

 

[Saitama]

This is one of the most sadistic questions I've seen. Where are you getting these questions?

From practice sheets. Do you want me to send the links to those practice sheets by PM? I don't know if its ok to post them here.

 

[verty]

From practice sheets. Do you want me to send the links to those practice sheets by PM? I don't know if its ok to post them here.

A PM will be fine, thanks. I haven't seen these type questions before, is all.